Question

(x^2 -4)^1/2dx的不定积分怎么求？

Answer 1

令 x = 2secu, 则
I = ∫2tanu 2secutanudu = 4∫secu(tanu)^2du = 4∫secu[(secu)^2-1]du
= 4∫(secu)^3du - 4∫secudu = 4∫secudtanu - 4∫secudu
= 4secutanu - 4∫secu(tanu)^2du - 4∫secudu
= 4secutanu - I - 4∫secudu
解得 I = 2secutanu - 2∫secudu = 2secutanu - 2ln|secu+tanu| + C
= (1/2)x√(x^2-4) - 2ln|x/2+√(x^2-4)/2| + C
= (1/2)x√(x^2-4) - 2ln|x+√(x^2-4)| + C1

Answer 2

let
x= 2secu
dx =2secu.tanu du
∫ (x^2-4)^(1/2) dx
=∫ (2tanu) [2secu.tanu du ]
=4∫ (tanu)^2.secu du
=4∫ tanu dsecu
=4tanu.secu -4∫ (secu)^3 du
=4tanu.secu -2[secu.tanu +ln|secu +tanu|] +C'
=2tanu.secu -2ln|secu +tanu| +C'
=2(1/2)√(x^2-4).(x/2) -2ln|(x/2) +(1/2)√(x^2-4)| +C'
=(1/2)x√(x^2-4) -2ln|x+√(x^2-4)| +C
//

∫ (secu)^3 du
=∫ secu dtanu
=secu.tanu -∫ secu.(tanu)^2 du
=secu.tanu -∫ secu.[(secu)^2-1] du
2∫ (secu)^3 du=secu.tanu +∫ secu du
∫ (secu)^3 du
=(1/2)[secu.tanu +ln|secu +tanu|] +C

Answer 3

详细过程如图，希望能帮到你解决你燃眉之急